cube files
Indexed Cube Lovers Archive
3-face subgroup of the 3x3x3 cube
Submitted by secondmouse on Fri, 05/07/2010 - 08:36.Somewhat surprisingly I was unable to find any tables for
the number of positions of various lengths in ascending order
for this order 170659735142400 group w.r.t. three mutually orthogonal face generators such as U,F and R, i.e.
This is seemingly beyond the reach of conventional software tools to solve, so I am appealing to the seemingly large number of contributors who have developed their own bespoke utilities. I'd be grateful if a table could be posted here. In particular is the diameter of the Cayley graph for this known?
free rubiks cubes
Submitted by whatsupdog1 on Thu, 05/06/2010 - 21:17.to get free cubes and more click on this banner
All the Syllables, Corners Group, Quarter Turn Metric
Submitted by Jerry Bryan on Thu, 04/22/2010 - 22:03.My recent posting on |EndsWith| values for the corners of the 3x3x3 in the quarter turn metric was really secondary to another project I was working on. Namely, I was working on determining all the syllables for the corners of the 3x3x3 in the quarter turn metric. I discovered many more syllables than I was expecting. As a part of determining why there were so many syllables, I discovered that the |EndsWith| values were much higher than I was expecting. To a certain extent, one can think of the higher |EndsWith| values as being the "cause" of there being a larger number of syllables than expected.
» 10 comments | read more
EndsWith Values, Corners Group, Quarter Turn Metric
Submitted by Jerry Bryan on Sun, 04/11/2010 - 11:19.My calculations of God's algorithm have generally included an analysis of the distribution of EndsWith values. My best God's algorithm results for the quarter turn metric are out to 13q. Tom Rokicki has since calculated out to 15q. Out to 13q, it is the case that for the vast majority of positions x we have |EndsWith(x)| = 1. Tom may be able to speak to the 14q and 15q cases, but I cannot.
Given the predominance of |EndsWith(x)| = 1 out to 13q, I have assumed that for the vast majority all of cube space it's probably the case that |EndsWith(x)| = 1. I no longer believe that assumption is correct. Which is to say, I now have an EndsWith distribution for the entirety of the corners group in the quarter turn metric. Beyond a certain distance from Start in the corners group, there are no instances of |EndsWith(x)| = 1 whatsoever, and overall the |EndsWith(x)| = 1 cases are very much in the minority. It now seems to me that the same is probably true for the complete cube group including corners and edges. I suspect the reason we have not seen this effect for the complete cube group is simply that we have not yet been able to calculate out far enough from Start.
» 9 comments | read more
algorithm for generating permutations for the rubik cube
Submitted by nooneimportant on Fri, 04/02/2010 - 11:13.hi,
let's say I want to distribute permutation checking over say 10 computers.
so if there are a total of n permutations the first machine will check n/10.
the second will check from n/10 to 2n/10
third will check from 2n/10 to 3n/10.
and so forth.
the algorithm that generates permtuations needs to generate the ith permutation in O(1)
so that I can efficiently start each machine's work.
what algos for generating permtuations do you know that can do this ?
thanks
let's say I want to distribute permutation checking over say 10 computers.
so if there are a total of n permutations the first machine will check n/10.
the second will check from n/10 to 2n/10
third will check from 2n/10 to 3n/10.
and so forth.
the algorithm that generates permtuations needs to generate the ith permutation in O(1)
so that I can efficiently start each machine's work.
what algos for generating permtuations do you know that can do this ?
thanks
Second largest coset solved yet of actual Rubik's positions
Submitted by mdlazreg on Fri, 04/02/2010 - 05:43.After solving the trivial coset [All positions that have the same orientation as the START position] published here Largest coset solved yet of actual Rubik's positions
I went ahead and calculated the full distribution of the flipped coset [All positions that have the same orientation as Reid's position (The only known 26q*)].
Enjoy:
I went ahead and calculated the full distribution of the flipped coset [All positions that have the same orientation as Reid's position (The only known 26q*)].
Enjoy:
Depth Trivial coset Flipped coset 0 1 0 1 4 0 2 10 0
» 4 comments | read more
basic algorithms and schreier sims ?
Submitted by nooneimportant on Sat, 03/27/2010 - 17:30.hi, where can I find a comprehensive exposition with explanations and examples, the kind of stuff that you can really learn from ?
I'm trying to implement this for solving the cube, it's a hobby project of mine.
I've noticed that another approach(if I use some basic algos like http://www.ryanheise.com/cube/beginner.html ) would be to pattern match the cube and "hardcode" all the cases of ryan heise and this would also yield a solution.
I'm a begginer with the cube, I almost solved it in reality and would like to write code to solve it.
(I have already set up an opengl simulation and
I'm trying to implement this for solving the cube, it's a hobby project of mine.
I've noticed that another approach(if I use some basic algos like http://www.ryanheise.com/cube/beginner.html ) would be to pattern match the cube and "hardcode" all the cases of ryan heise and this would also yield a solution.
I'm a begginer with the cube, I almost solved it in reality and would like to write code to solve it.
(I have already set up an opengl simulation and
» 4 comments | read more
UF and RBL generate the whole cube group
Submitted by rokicki on Sat, 03/20/2010 - 13:43.The two simple generators UF and RBL (without any rotations) generate the whole cube group.
It's surprising to me that these two very simple generators suffice.
It's easy to see no shorter set of generators (expressed in face turns) suffice because you need at least five faces.
It's surprising to me that these two very simple generators suffice.
It's easy to see no shorter set of generators (expressed in face turns) suffice because you need at least five faces.
Presentation for Rubik's cube
Submitted by jaap on Fri, 03/12/2010 - 03:32.I just found a recent post by "secondmouse" on sci.math that deserves a wider audience. I'll quote it here in full.
I found the following short presentation for the
miniature 2x2x2 Rubik's cube of order 3674160:
< a,b,c | a^4 = b^4 = c^4 = 1,
ababa = babab,
bcbcb = cbcbc,
abcba = bcbac,
bcacb = cacba,
cabac = abacb,
(ac)^2 (ab)^3 (cb)^2 = 1 >
See the following link for more info as to why» 15 comments | read more
1,000,000 cubes optimally solved in both QTM and FTM
Submitted by rokicki on Sun, 03/07/2010 - 19:43.I have solved all 1,000,000 random cube positions from the earlier
article now in both QTM and HTM. Here is the resulting grid:
12h 13h 14h 15h 16h 17h 18h 19h sum
15q 1 1 3 2 - - - - 7
16q - 2 18 48 35 - - - 103
17q - 3 23 143 347 354 - - 870
18q - 5 40 305 1713 4520 2034 - 8617
19q - 1 40 505 5190 29711 33363 474 69284
20q - 2 39 674 9932 100164 212466 7213 330490
21q - - 9 345 7697 104052 301668 16371 430142
22q - - - 41 1533 28173 120449 9720 159916» 6 comments | read more
